−5 1/5÷(2/3) help plz

Justin has $500 in a savings account at the beginning of the summer. He wants to have at least $300 in the account by the end of

SSSSS [86.1K]

Answer:

10 weeks

Step-by-step explanation:

1 because 20 times 10 is 200 and if he starts with 500 then he can withdraw 20 dollars a week for 10 weeks to have 300 left by the end of the summer, hope this helps :)

3 0

3 years ago

Does anyone have a discord server for homework help?

Deffense [45]

Answer:

I do....................................

7 0

2 years ago

A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean

elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}$ ). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0

3 years ago

1. (a). A police car, approaching right-angled intersection from the north, is chasing a speedmg

tamaranim1 [39]

Answer:

<em>The SUV is running at 70 km/h</em>

Step-by-step explanation:

<u>Speed As Rate Of Change </u>

The speed can be understood as the rate of change of the distance in time. When the distance increases with time, the speed is positive and vice-versa. The instantaneous rate of change of the distance allows us to find the speed as a function of time.

This is the situation. A police car is 0.6 Km above the intersection and is approaching it at 60 km/h. Since the distance is decreasing, this speed is negative. On the other side, the SUV is 0.8 km east of intersection running from the police. The distance is increasing, so the speed should be positive. The distance traveled by the police car (y) and the distance traveled by the SUV (x) form a right triangle whose hypotenuse is the distance between them (d). We have:

$d=\sqrt{x^2+y^2}$

To find the instant speeds, we need to compute the derivative of d respect to the time (t). Since d,x, and y depend on time, we apply the chain rule as follows:

$\displaystyle d\ '=\frac{x.x'+y.y'}{\sqrt{x^2+y^2}}$